Review of Classical Statistical Estimation
1. Parameters of probability distribution
e.g. , , ,  are constants (unknown)
2.
Observed data
Choice of
probability
distribution
Point
estimates of
parameter
Computation
of probability
3. Statistical uncertainty of estimation
described in terms of confidence intervals
not incorporated in probability computation
(same applies to model error)
4. Subjective judgmental information not
incorporated in estimation
5. Large error may incur if sample size is
small
Accident Rate
at an Intersection
Theoretical Model
Previous Experiences
Computer Simulation
Observed performance
first week

1 per
year
1 in
=?
mean rate of accident
EXAMPLE – SHORT PILE
p = fraction of short piles
1. Engineer’s Experience & Judgement
P’(p = pi)
.4
.3
.15
0
.2
.4
.6
.1
.05
.8
1.0
pi
p̂ = mean of p = 0.44
2. After 1 successful detection test,
i.e. pile selected for inspection is found to be short
P”(p = 0.2) = P(p = 0.2 | successful test)
P(suc.test | p = 0.2)P(p = 0.2)
= P(suc.test | p = 0.2)P(p = 0.2)+...+P(suc.test | p = 1.0)P(p = 1.0)
0.2  0.3
= 0.2  0.3  0.4  0.4  0.6  0.15  0.8  0.1  1.0  0.05
= 0.136
P”(p = pi)
.364
p̂ = 0.55
.204 .182
.136
0
.2
.4
.6
.8
.114
1.0
pi
P”(p = pi)
P”(p = pi)
P”(p = pi)
0.5
0.5
0.5
n=2
0.2
0.4
0.6
n=6
n=4
0.8
pi
1.0
P”(p = pi)
0.2
0.4
0.6
0.8
1.0
P”(p = pi)
pi
0.2
n = 10
0.5
0.6
0.8
pi
1.0
Fig. E8.1c
0.2
0.4
0.6
0.8
1.0
pi
0.2
0.4
Case of no specific prior information
 diffused prior distribution
.2
.2
.2
Prior PMF
.2
Prior PDF
p
0
1.0
pi
0.5
0.6
PMF of p for increasing number of successful tests
.2
0.8
n=
0.5
0.4
0.6
P”(p = pi)
n=8
0.2
0.4
1
0.8
1.0
pi
Incorporation of Statistical Uncertainty into
Probability Calculation
Problem:
Suppose three piles are selected at random
to be inspected in a new but similar site,
what is the probability that none of these 3
piles will be short? (say, after 1 succ. test)
1. Classical method: No judgement
use point estimate p̂ = 1.0
 3 0 3
P(X  0)    1 0  0
 0
2. Semi-Bayesian: Include judgment
use point estimate p̂ = 0.55
 3
P(X = 0) =  0  (0.55)0(0.45)3 = 0.09
 
3. Bayesian Method: Include Judgement,
use entire distribution of p
(through theorem of total probability)
P(X = 0)
= P(X=0 | p=0.2)P”(p=0.2) +…+P(X=0 | p=1)P”(p=1)
= (0.8)3  0.136 + (0.6)3  0.364
+ (0.4)3  0.204 + (0.2)3  0.182
= 0.163
In short, Bayesian
approach
1) allows judgmental
information to be
incorporated
2) permits updating of
probability and distribution
of parameter
3) combines the statistical
uncertainty of parameter
with basic variability of
the random variable
Let  be the parameter
Posterior
Prior
P”( = i)
P( |    i ) P '(   i )
 P (   i |  ) 
 P( |    i ) P '(   i )
i
observed
data
P( X  a)   P( X  a |    i ) P "(   i )
i
Bayesian probability
Continuous Case
Prior
f’()
Posterior
f"()

i
f "( i ) 
P”( = i)
i + i
P( |  i ) f '( i )
 P( |    i ) f '( i )
i
P( |  ) f '( )
as   0, f "( ) 

 P( |  ) f '( )d

In short

Normalizing constant such that
 f "( )d  1.0

f"() = k L() f’()
Likelihood function
P( | ) i.e. given value 
what is probability of observing data


k    L( ) f ' ( )d 
 


P( X  a) 
1
 P( X  a |  ) f "( )d
